⎧
1+x 2,x ≤0,e -x ,x >0,
则⎠⎛13f (x -2)d x =( )
A .3+ 1
e
B .2-e C.73- 1 e
D .2-
1 e
解析:选C ⎠⎛1
3f (x -2)d x =⎠⎛1
2f (x -2)d x +⎠⎛2
3f (x -2)d x =⎠⎛1
2(x 2-4x +5)d x +⎠
⎛2
3e
-x +2
d x
=⎣⎡⎦⎤13x 3
-2x 2+5x |
2
1+(-e
-x +2)|
3
2=
⎣⎡⎦
⎤⎣⎡⎦⎤13×23-2×22+5×2-⎣⎡⎦⎤13×13-2×12+5×1+[(-e
-3+2
)-(-e
-2+2
)]=73
-1
e ,故选C.
4.(2018·吉林长春调研)若f (x )=x 2+2⎠⎛01f (x )d x ,则⎠⎛0
1f (x )d x =( )
A .-1
B .-13
C.13
D .1
解析:选B 设⎠⎛0
1f (x )d x =c ,则f (x )=x 2+2c ,所以⎠
⎛0
1f (x )d x =1
3x 3
|
1
0+2cx
|
1
0=13
+2c =c ,解得c =-1
3
,故选B.
5.(2018·山东陵县一中月考)定积分⎠
⎛0
1x -1
3d x 的值为________.
解析:⎠
⎛0
1x -13d x =32x
2
3| 10
=32-0=3
2
. 答案:3
2
6.(2018·安徽蚌埠摸底)⎠
⎛1-1(|x |+s in x )d x =________.
解析:⎠⎛1-1(|x |+s in x )d x =⎠⎛1-1|x |d x +⎠
⎛1-1s in x d x .根据定积分的几何意义可知,函
数y =|x |在[-1,1]上的图象与x 轴,直线x =-1,x =1围成的平面区域的面积为1.y =s in x 为奇函数,则⎠⎛1-1s in x d x =0,所以⎠
⎛1-1(|x |+s in x )d x =1.
答案:1
对点练(二) 定积分的应用
1.(2018·福建南平期中)两曲线y =s in x ,y =cos x 与两直线x =0,x =π2所围成的平面
区域的面积为( )
解析:选D 作出曲线y =s in x ,y =cos x 与两直线x =0,x =π
2 所围成的平面区域,
如图.
根据对称性可知,曲线y =s in x ,y =cos x 与两直线x =0,x =π
2 所围成的平面区域的
面积为曲线y =s in x ,y =cos x 与直线x =0,x =π
4所围成的平面区域的面积的两倍,所以S
=
(cos x -s in x )d x .故选D.
2.(2018·武汉模拟)设变力F(x )(单位:N )作用在质点M 上,使M 沿x 轴正方向从x =
1 m 处运动到x =10 m 处,已知F(x )=x 2+1且方向和x 轴正方向相同,则变力F(x )对质点M 所做的功为( )
A .1 J
B .10 J
C .342 J
D .432 J
解析:选C 变力F(x )=x 2+1.使质点M 沿x 轴正方向从x =1运动到x =10所做的功W =∫101F(x )d x =∫101
(x 2+1)d x =⎝⎛⎭
⎫13x 3+x |
10
1=342(J).
3.(2018·广东七校联考)由曲线xy =1,直线y =x ,y =3所围成的平面图形的面积为( ) A.329 B .2-l n 3 C .4+l n 3
D .4-l n 3
解析:选D S =⎠⎛113⎣⎡⎦⎤3-1x d x +1
2×2×2=(3x -l n x )| 11
3+2=4-
l n 3,故选D.
4.(2018·河南安阳调研)由曲线y =2x ,直线y =x -3及x 轴所围成的图形的面积为( )
A .12
B .24
C .16
D .18
解析:选D 曲线y =2x ,直线y =x -3的交点为(9,6),由定积分的几何意义可知,曲线y =2x 与直线y =x -3及x 轴围成的面积为⎠
⎛0
9[2x -(x -3)]d x -1
2×3×3=
⎝⎛⎭
⎫
43x 32-12x 2+3x |
9
-92
=18,故选D.
5.(2018·福建省师大附中等校期中)已知函数f (x )=-x 3+ax 2+bx (a ,b ∈R)的图象如图所示,它与x 轴相切于原点,且x 轴与函数图象所围成区域(图中阴影部分)的面积为1
12
,则a 的值为( )
A .0
B .1
C .-1
D .-2
解析:选C f ′(x )=-3x 2+2ax +b .由题意得f ′(0)=0,得b =0,∴f (x )=-x 2(x -a ).由
⎠
⎛a
0(x 3
-ax 2
)d x =⎝⎛⎭⎫14x 4-13a x 3|
a =0-a 44
+a 43
=
a 412
=1
12
,得a =±1.函数f (x )与x 轴的交点的横坐标一个为0,另一个为a .根据图形可知a <0,即a =-1.
6.从空中自由下落的一物体,在第一秒末恰经过电视塔顶,在第二秒末物体落地,已知自由落体的运动速度为v =g t (g 为常数),则电视塔高为( )
A.12
g B .g